# Covariant/contravariant transform and metric tensor

1. Sep 7, 2012

### nigelscott

H is a contravariant transformation matrix, M is a covariant transformation matrix, G is the metric tensor and G-1 is its inverse. Consider an oblique coordinates system with angle between the axes = α

I have G = 1/sin2α{(1 -cosα),(-cosα 1)} <- 2 x 2 matrix

I compute H = G*M where M = {(1 0), (cosα sinα)} and get H = {(1 -1/tanα),(0 1/sina)} which is what I expect.

Now I want go from H back to M so I compute M = G-1H

So by my reckoning G-1 = 1/sin4α{(1 cosα),(cosα 1)}

But when I multiply G-1 and H I don't get back to M. The is a 1/sin4α multiplying the whole thing.
What am I missing?

2. Sep 7, 2012

### Muphrid

When you say G is the metric tensor, the metric tensor converts contravariant to covariant? Or is it the other way around?

Actually, I think it's pretty clear that the issue is with your computed $G^{-1}$. The operator $G$ has determinant 1. You've got an extra factor of $1/\sin^2 \alpha$ for some reason. Did you forget the prefactor from $G$ somewhere?

Last edited: Sep 7, 2012
3. Sep 8, 2012

### nigelscott

contravariant = G * covariant
covariant = G-1 * contravariant

If G = 1/sin2α{(1 -cosα),(-cosα 1)}

then wouldn't G-1 = (1/sin2α)(1/|detG|){(1 +cosα),(+cosα 1)}

which gives 1/sin4α{(1 +cosα),(+cosα 1)}

Now if I compute GG-1 I get 1/sin4{(sin2α 0),(0 sin2α)} which is equivalent to 1/sin2α{(1 0),(0 1)}

So I get the identity matrix multiplied by 1/sin2α which doesn't add up.

Incidentally, this question came up as a result of watching this video http://www.youtube.com/watch?v=qDLmJE2bOy4

4. Sep 8, 2012

### Muphrid

You're saying $\det G = 1/\sin^2 \alpha$? Because it's not.

5. Sep 8, 2012

### nigelscott

No, I thought I had 1/|detG| where |detG| = 1 - cos2α = sin2α

6. Sep 8, 2012

### Muphrid

It's not $\sin^2 \alpha$ either. Why don't you go through the whole calculation of the determinant?

7. Sep 8, 2012

### nigelscott

G = 1/sin2a{(1 -cosa),(-cosa 1)} = {(1/sin2a -cosa/sin2a),(-cosa/sin2a 1/sin2a)}
detG = 1/sin4a - cos2a/sin4a = 1/sin2a
Thus, 1/|detG| = sin2a
So G-1 = sin2a{(1 +cosa),(+cosa 1)}
Now GG-1 does equal the identity matrix. Am I good so far? If so I will return to my original question in the next post.

8. Sep 8, 2012

### Muphrid

Okay, now that you're getting the identity for $GG^{-1}$, I expect the problems you were having will be resolved.

9. Sep 9, 2012

### nigelscott

Yes, I got it to work when I used the form G-1 = sin2α(1 +cosα),(+cosα 1)}. If I use G-1 = {(sin2α +cosα.sin2α),(cosα.sin2α,1)} it doesn't.